Question

A \(1.604-\mathrm{g}\) sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(6.400 \mathrm{~g}\) oxygen gas are sealed into a 2.50-L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is \(0.326 \mathrm{~atm}\), and the pressure of water vapor is \(4.45\) atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

Short answer

The equilibrium pressures of carbon monoxide and carbon dioxide are both 1.0385 atm.

Step-by-step solution

Find moles of reactants and products

First, we need to find how many moles of each reactant are present initially. We can use the initial mass given and the molar mass of the substances. For methane (CH4), the molar mass is: \(16.04 \dfrac{\mathrm{g}}{\mathrm{mol}}\) (12.01 g/mol for C and 4.03 g/mol for 4 x H) For oxygen (O2), the molar mass is: \(32.00 \dfrac{\mathrm{g}}{\mathrm{mol}}\) (16 g/mol for 2 x O) Using these molar masses, we can find the initial moles of methane and oxygen: initial moles of methane: \(\dfrac{1.604\,\text{g}}{16.04 \,\text{g/mol}} = 0.1000 \,\text{mol}\) initial moles of oxygen: \(\dfrac{6.400\,\text{g}}{32.00 \,\text{g/mol}} = 0.2000 \,\text{mol}\)

Use the Ideal Gas Law to find initial pressures

Now we will use the ideal gas law to find the initial pressures of methane and oxygen. The ideal gas law is given by: \(PV = nRT\) where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin. The value of R used for pressure in atm is: \(R = 0.0821 \dfrac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}}\) First, convert the temperature to Kelvin: \(T = 411 + 273.15 = 684.15\, \text{K}\) Now, we can find the initial pressures: Initial pressure of methane: \(P_{CH4} = \dfrac{n_{CH4}RT}{V} = \dfrac{0.1000 \,\text{mol} \cdot 0.0821 \,\dfrac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}} \cdot 684.15\, \text{K}}{2.50\,\text{L}} = 2.240\, \text{atm}\) Initial pressure of oxygen: \(P_{O2} = \dfrac{n_{O2}RT}{V} = \dfrac{0.2000 \,\text{mol} \cdot 0.0821 \,\dfrac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}} \cdot 684.15\, \text{K}}{2.50\,\text{L}} = 4.48\, \text{atm}\)

Determine the change in pressures during the reaction

At equilibrium, the pressure of oxygen is 0.326 atm. The initial pressure of oxygen was 4.48 atm. Therefore, the pressure of oxygen has decreased by: \(\Delta P_{O2} = 4.48 - 0.326 = 4.154\, \text{atm}\) Since we know the initial pressure of methane was 2.240 atm and we can assume that all methane has reacted. Also, considering the stoichiometry of the reaction, 1 mole of CH4 reacts with 2 moles of O2, we can find out the ΔP for CH4 as: \(\Delta P_{CH4} = \dfrac{1}{2} \times \Delta P_{O2} = \dfrac{1}{2} \times 4.154 = 2.077\, \text{atm}\)

Find the final equilibrium pressures

Now, since we know that \(\Delta P_{CH4} = 2.077\) and its start pressure is 2.24 atm, it indicates that all the methane has been consumed. This change also produces a final equilibrium pressure of water vapor of 4.45 atm. Using stoichiometry of the reaction, we can say that the change in pressure of CH4 will be equal to the sum of changes in pressure of CO and CO2: \(\Delta P_{CO} + \Delta P_{CO2} = \Delta P_{CH4} = 2.077\, \text{atm}\) We also know that since 2 moles of O2 reacts with 1 mole of CH4, the change in pressure of CO and CO2 would be equal: \(\Delta P_{CO} = \Delta P_{CO2}\) Thus, we can say that: \(2\Delta P_{CO} = 2.077\, \text{atm}\) \(\Delta P_{CO} = 1.0385\, \text{atm}\) Finally, we can find the equilibrium pressures of CO and CO2: Final equilibrium pressure of CO: \(P_{CO} = P_{CH4} + \Delta P_{CO} = 0 + 1.0385 = 1.0385\, \text{atm}\) Final equilibrium pressure of CO2: \(P_{CO2} = P_{CH4} + \Delta P_{CO2} = 0 + 1.0385 = 1.0385\, \text{atm}\) The equilibrium pressures of carbon monoxide and carbon dioxide are both 1.0385 atm.

Key concepts

Understanding the Ideal Gas Law

The Ideal Gas Law is a pivotal equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. It's expressed as \(PV = nRT\), with P representing pressure, V standing for volume, n indicating moles of gas, R being the ideal gas constant, and T the temperature in Kelvin. In our exercise, we exploited the Ideal Gas Law to determine the initial pressures of methane and oxygen before the reaction takes place.

Temperature, always used in Kelvin for gas law calculations, accounts for the absolute scale directly tied to the energy present in the gas particles. Converting Celsius to Kelvin is therefore critical. For pressure in atmospheric units (atm), R is valued at \(0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}}\). This value enables us to comprehend the state of our reactants and anticipate how they behave under changing conditions, setting the stage for understanding chemical equilibrium.

Reactants and Products in Equilibrium

In a chemical reaction reaching equilibrium, the concentrations of reactants and products remain constant over time. However, this does not mean their quantities are equal, but simply that their rates of formation and consumption balance out. In the given problem, we focus on the pressures instead of concentrations, as we're dealing with gases.

At equilibrium, the pressure of each gas is a direct reflection of its quantity within the reaction vessel. The exercise task was to determine the pressures of carbon monoxide (CO) and carbon dioxide (CO2) in a sealed vessel based on the known pressure of oxygen (O2) and water vapor (H2O). By understanding the stoichiometry and monitoring the pressure changes, we dissected how the system reached equilibrium and calculated the pressures of all species at this state.

The Role of Stoichiometry in Equilibrium

Stoichiometry is the math behind chemistry; it's how we balance equations and quantify the relationships between reactants and products. It involves mole ratios derived from a balanced chemical equation and is fundamental in predicting how much of each substance is needed or produced in a reaction.

When discussing equilibrium, we employ stoichiometry to interpret the shifts and interactions between gases involved. In our problem, we examined the stoichiometry of methane's reaction with oxygen. Each mole of methane requires two moles of oxygen to react; this 1:2 ratio becomes key to deducing the changes in pressure for each gas as the system reaches equilibrium. Using these ratios effectively enabled us to solve for the unknown pressures of carbon monoxide and carbon dioxide, illustrating the intrinsic ties between stoichiometry and equilibrium in chemical reactions.